8.5)
a)
El estimador que pusimos fue
$$\widehat{\theta}^* = \frac{\widehat{\theta}-b}{a}\\ \\ MSE(\widehat{\theta}^*) = V(\widehat{\theta}^*)+[B(\widehat{\theta}^*)]^2\\ \\ \text {Como era insesgado }\implies B(\widehat{\theta}^*)=0\\ \\ \\ MSE(\widehat{\theta}^*) = V(\widehat{\theta}^*)=\\ \\ \\ V \left(\frac {\widehat{\theta}-b}{a} \right) = \frac {1}{a^2} V(\widehat{\theta}-b)\\ \\ \text{Calculemos aparte } V(\widehat{\theta}-b)\\ \\ V(\widehat{\theta}-b)= \\ \\ E[(\widehat{\theta}-b)^2]-[E(\widehat{\theta}-b)]^2 =\\ \\ E(\widehat{\theta}^2)+b^2-2bE(\widehat{\theta})-[E(\widehat{\theta})-b]^2 =\\ \\ E(\widehat{\theta}^2)+b^2-2bE(\widehat{\theta})-[E(\widehat{\theta})]^2-b^2 +2bE(\widehat{\theta})=\\ \\ \\ E(\widehat{\theta}^2)-[E(\widehat{\theta})]^2 = V(\widehat{\theta})\\ \\ Luego\\ \\ MSE(\widehat{\theta}^*)= \frac{V(\widehat{\theta})}{a^2}\\ \\$$
b) Tenemos
$$MSE(\widehat{\theta}^*)=\frac{V(\widehat{\theta})}{a^2}\\ \\ \\ MSE(\widehat{\theta})= V(\widehat{\theta})+ [B(\widehat{\theta})]^2\\ \\ \text{Para cualquier valor a > 1 se cumplirá}\\ \\ MSE(\widehat{\theta}^*) \lt MSE(\widehat{\theta})\\ \\ \text{Como nos piden un ejemplo tomemos a=2}\\$$
c) Tenemos
$$\begin{align}&MSE(\widehat{\theta}^*)=\frac{V(\widehat{\theta})}{a^2}\\ &\\ &\\ &MSE(\widehat{\theta})= V(\widehat{\theta})+ [B(\widehat{\theta})]^2=\\ &V(\widehat{\theta})+[E(\widehat\theta)-\theta]^2=\\ &V(\widehat{\theta})+[a\theta+b-\theta]^2=\\ &V(\widehat{\theta})+[(a-1)\theta+b]^2\\ &\\ &\text{bastante lío hay, haremos b=0 para facilitar}\\ &\\ &\text{debemos hacer}\\ &\\ &\frac{V(\widehat{\theta})}{a^2} \gt V(\widehat{\theta})+(a-1)^2\theta^2\\ &\\ &\text {Si tomamos } 0\le a \lt 1 \text { el último sumando está acotado}\\ &\text{hagamos que lo cumpla para el peor de los casos}\\ &\\ &\\ &\frac{V(\widehat{\theta})}{a^2} \gt V(\widehat{\theta})+\theta^2\\ &\\ &\\ &a < \sqrt{\frac{V(\widehat{\theta})}{V(\widehat{\theta})+\theta^2}}\end{align}$$
Tomando un a que cumpla eso y b=0 se cumplirá la desigualdad.
Y eso es todo.